That's a reasonable answer. I think my favorite way of viewing this question is to think of a complete graph on six vertices (i.e. each vertex is connected to each other vertex), where all edges are colored either red or blue. Then you are to show that there is either a red triangle or a blue triangle.
You could put your reasoning on this picture, as it becomes very easy to follow.
This is also a good time to first learn about Ramsey Theory - of which this is one of the easiest examples. Look here.
Let $G(V,E)$ be a $4$-regular simple graph on $17$ vertices. We claim that there are two vertices $u,v\in V$ such that $u\neq v$, $u$ is not adjacent to $v$, and $u$ and $v$ do not have a common neighbor. We prove by contradiction by assuming that, for any two distinct vertices $u$ and $v$ of $G$, if $u$ is not adjacent to $v$, then $u$ and $v$ have a common neighbor.
Let $S$ denote all pairs $\big(\{u,v\},w\big)$ with $u,v,w\in V$ such that $u\neq v$, $\{u,v\}\notin E$, and $w$ is adjacent to both $u$ and $v$. For each $w\in V$, $w$ has four neighbors. Therefore, at most $\binom{4}{2}$ pairs of neighbors of $w$ are not adjacent. This proves that
$$|S|\leq \binom{4}{2}\,|V|=6\,|V|=102\,.\tag{*}$$
Now, $|E|=\dfrac{4\cdot |V|}{2}=2\,|V|=34$ by the Handshake Lemma. Thus, there are $$\binom{17}{2}-|E|=102$$ pairs of vertices $\{u,v\}$ that are not edges of $G$. Each anti-edge pair $\{u,v\}$ produces at least one element of $S$, due to our hypothesis on $G$. This proves that $$|S|\geq 102\,.\tag{#}$$
From (*) and (#), we must have $|S|=102$. For (#) to be an equality, every anti-edge pair $\{u,v\}$ must have exactly one common neighbor $w\in V$. Additionally, $G$ must be a triangle-free graph for (*) to become an equality. This means $G$ is both triangle-free and quadrilateral-free. Therefore, $G$ is a graph on $17=4^2+1$ vertices with girth $g\geq 5$ in which all vertices have degree $4$. By the Hoffman-Singleton Theorem (for a proof, see here), if there exists an $r$-regular simple graph on $r^2+1$ vertices with girth at least $5$, then $r\in\{1,2,3,7,57\}$ (we know a partial converse, that is, for $r\in\{1,2,3,7\}$, there exists such a graph, but it is still a mystery for $r=57$, as you may guess, it is not easy to construct a graph on $57^2+1=3250$ vertices). This yields the desired contradiction.
Best Answer
Consider $6$ people represented by dots in a circle. Let every dot be connected to every other dot by a line and let the lines be the color red if the two people know each other and blue if they do not know each other. Consider any of the $6$ dots, say dot $x$. We can see that $x$ is connected to $5$ other dots by a line and by the pigeonhole principle $3$ of these lines are the same color, say red (the proof is analogous if we choose blue instead of red) . Now examine the $3$ dots connected to $x$. If any of those $3$ people are connected by a red line, then we have found a red triangle which represents $3$ people who know each other. If the $3$ people are not connected by any red lines, then all $3$ of them are connected by blue lines forming a blue triangle which represents $3$ people not knowing each other. Thus in a group of $6$ people there will always be $3$ people who know each other or $3$ people who do not know each other.
The pigeonhole principle applies because every person is related to every other person in two ways, (either they know the person or they do not). So when we consider any of the $6$ people we know that they know $3$ people or they don't know $3$ people.